Question

将分式(x²+3x)/(x+1)(x²+1)化为部分分式)

Answer 1

(x²+3x)/[(x+1)(x²+1)] = a/(x+1) + (bx+c)/(x^2+1)
= [(x+1)(bx+c)+a(x^2+1)]/[(x+1)(x²+1)]
b+a = 1, b+c = 3, c+a = 0,
a = -1, b = 2, c = 1,
(x²+3x)/[(x+1)(x²+1)] = -1/(x+1) + (2x+1)/(x^2+1)

Answer 2

设(x²+3x)/[(x+1)(x²+1)]=a/(x+1)+(bx+c)/(x^2+1),
去分母得x^2+3x=a(x^2+1)+(bx+c)(x+1)
=(a+b)x^2+(b+c)x+a+c,
比较系数得a+b=1,b+c=3,a+c=0,
解得a=-1,b=2,c=1.
所以(x²+3x)/[(x+1)(x²+1)]=-1/(x+1)+(2x+1)/(x^2+1)

Answer 3

解:设原式=(x²+1+3x-1)/(x+1)(x²+1)
=(x²+1)/(x+1)(x²+1)+(3x-1)/(x+1)(x²+1)
=1/(x+1)+(3x+3-4)/(x+1)(x²+1)
=1/(x+1)+3/(x²+1)-4/(x+1)(x²+1)

Answer 4

设(x²+3x)/[(x+1)(x²+1)]=a/(x+1)+(bx+c)/(x²+1)

=[a(x²+1)+(x+1)(bx+c)]/[(x+1)(x²+1)](通分)

x²+3x=a(x²+1)+(x+1)(bx+c)（只看分子）

令x=-1(别管原式分母=0，上面一行是整式，随意令，越简单越好)，-2=2a，a=-1

x²+3x=-(x²+1)+(x+1)(bx+c)

令x=0，0=-1+c，c=1

x²+3x=-(x²+1)+(x+1)(bx+1)，

令x=1，4=-2+2(b+1)，b=2

(x²+3x)/[(x+1)(x²+1)]=(2x+1)/(x²+1)-1/(x+1)

另一种方法：把第3行右边乘出来合并同类项，与左边比较系数可得a、b、c